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Math Help - A Trig Application

  1. #1
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    A Trig Application

    I have this question to do for my next class:

    A gutter is to be made from a metal sheet of width 3 metres by bending up one third of the sheet on each side through an angle θ. How should the θ be chosen so that the gutter will carry the maximum amount of water?

    This is what I pictured it to be:

    However, after that, I can't figure out where to start.

    Could someone help me out? A full solution would be very much appreciated.
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  2. #2
    GAMMA Mathematics
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    I wish I could show you a drawing, but here goes:

    \ | | /
    \|___|/

    The angle sides and the bottom are all of length one. We want to maximize area which we can split into three parts:

    Triangular cross sectional areas on the sides of the gutter, and the middle square c-s area. For some angle, \theta, the height is cos(\theta) and the top "length" is sin(\theta). The area of such triangle is 0.5sin(\theta)cos(\theta). The area of the square is sin(\theta). Add up the areas to get:

    sin(\theta)cos(\theta)+sin(\theta)

    To maximize the area, we need to take the derivative w.r.t. theta and set it equal to zero.

    cos^2(\theta)-sin^2(\theta)-sin(\theta) = 0

    \theta=\frac{\pi}{6}
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  3. #3
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    Oh. So would that be the angle formed in the triangle cross sectional area?

    Am I looking for the angle of that entire angle though? So would my final answer be 2pi/3?
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  4. #4
    GAMMA Mathematics
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    Quote Originally Posted by ty2391 View Post
    Oh. So would that be the angle formed in the triangle cross sectional area?

    Am I looking for the angle of that entire angle though? So would my final answer be 2pi/3?
    Yes, that angle is for the triangle, so you are right that you would have to add \frac{\pi}{2} to get an angle of: \frac{2\pi}{3}.
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